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Problem 901

AMC 12 late, AIME early

Geometry Difficulty 4.7 Prove it All-Soviet-Union Mathematical Olympiad · Soviet Union

Given 100 points on the plane. Prove that you can cover them with a collection of circles whose diameters total less than 100 and the distance between any two of which is more than 1. [The distance between circles radii $r$ and $s$ with centers a distance $d$ apart is the greater of 0 and $d - r - s$ .]

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

If we have two circles diameters $d$ and $d'$ , the distance between which is less than $1$ , then they are contained in a circle diameter $d + d' + 1$ . [If the line through the centers cuts the circles in $A$ , $B$ , $A'$ , $B'$ , then take a circle diameter $AB'$ .] So start with $100$ circles of diameter $1/1000$ each. If any pair is a distance $\leq 1$ apart, then replace them by a single circle, increasing the total diameter by $1$ . Repeat until all the circles are a distance $>1$ apart. We must end up with at least one circle, so the total increase is at most $99$ . Hence the final total diameter is at most $99\,1/10$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.